Explanation of Michelson Interferometer Fringe Shift I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides. 
 A: This is speculating - but if your slides are of non-uniform thickness, or they are bent as a result of the pinching, they will present a different path length in one leg of the interferometer (and therefore give rise to a shift in the fringe pattern). This may become clear by looking at this diagram:

In the diagram on the left, the total path length is independent of the position of the ray - in all cases the light bends by the same amount as it interacts with the different surfaces. In the diagram on the right, the rays closer to the "pinch point" will traverse less glass than the ones that are further away (which intersect the glass at a greater angle). This means that the path length will change as you move the slides left to right.
It is not clear whether there is a spacer as part of your "wedge" (I imagine there must be one, but I can't see it in your photo). If there is, then the big clip you use will surely bend the slides; and a Michelson interferometer is very, very sensitive to path length differences...
A: At a guess, the effect rises from the fact that your interferometer is not properly aligned. The presence of linear, rather than circular, fringes suggests that there is an angular misalignment. Then moving the wedge causes a lateral shift in the intersection point of the beam and the angled slide, which results in a shift in the apparent position of the beam at the target.
Try aligning the system to produce a bull's eye fringe pattern, and see if the anomaly persists. 
A: This answer is based on Floris' insight that the slides might be bent.
Let's say the laser hits the slide at an angle of $\theta$ and travels through the panel at an angle of $\theta'=\sin^{-1}({n_a\over n_s}\sin(\theta))$. Let's assume the curvature is light enough that the laser essentially exits parallel to how it entered. I am also going to assume you are using a $\lambda=633\mathrm{nm}$ laser. We can express the change in phase from having no slides as $\phi={4\pi D\over\lambda}\sec\theta'\left({n_s\over n_a}-\sec(\theta'-\theta)\right)$ using a little trigonometry, where $D$ is the thickness of a slide. We want the curvature which is ${d\theta\over ds}$, where $ds=\sec\theta\,dx$. Let's express ${d\theta\over ds} = \left({d\phi\over d\theta}\right)^{-1} \left({d\phi\over dx}\right) \cos\theta$. You measured $d\phi\over dx$ to be ${2\pi\over 4\mathrm{mm}}$, and specified $n_s=1.52={n_s\over n_a}$. Let's say our angle of incidence is 10 degrees. We will now give this mess to Wolfram.
So at this point, we have to have a curvature of .06 degrees per mm to observe this fringe effect.
A: In the 2 glasses there are 4 surfaces, i.e interface air/glass, and 8 surface orientations (a..h) and plenty room for interference between reflections and the main beam. 
LASER  (air)    a1b (glass) c2d      (air)       e3f  (glass)   g4h
At each interface the is reflection that will be reflected forward again (self-interference) 
look for iridiscence in thin materials.  
previous answer, not important now: 
Evaluate the dimensions of the screw thread that moves the clamp (distance, pitch) and how many turns are required for a given deviation.
I do not see the screw, but I imagine that it exists and influences the outcome.
